MATHEMATICS SAMPLE PAPER # 1 ANSWER AND SOLUTIONS

CLASS - X (CBSE) STANDARD

MATHEMATICS

MATHEMATICS

SAMPLE PAPER # 1

ANSWER AND SOLUTIONS

SECTION-A

4.

Volume of hemisphere =

2 3p

? 21 ?3 ?? 2 ??

1. 17th term from end means 17th term from begining of 216, 211, ..........., 6, 1 i.e. d = 211 ? 216 = ? 5

Volume of one cone =

1 3

p

? ??

7 2

?2 ??

?

9

a17 = a + 16d = 216 + 16(? 5) = 216 ? 80 = 136 2. 4x2 + 4 3x + 3 = 0

Here a = 4, b = 4 3, c = 3

2? 21 ? 21 ? 21 222

Number of cones = 7 ? 7 ?9 = 21 22

D = b2 ? 4ac

OR

3.

ALLEN D= (4 3)2 -4(4)(3)=48?48 =0

? Roots are Real and equal

Let AB and CD be two parallel tangents to the circle at P and Q respectively.

To prove : POQ is a diameter.

A

P

B

R

O

C

Q

D

Construction : Through O draw OR||BA or

OR || CD as AB and CD are parallel 5.

tangents.

Proof : ?OPA = 90? (radius is always

perpendicular to tangent)

Volume of cuboid = [66 ? 20 ? 27] cm3

= Volume of the pipe

= p[R2 ? r2]l

? p[R2 ? r2]l = 66 ? 20 ? 27

Here R = 5 cm and r = 4 cm

22 7

?

[52

?

42]l

=

66

?

20

?

27

66?20 ?27? 7

? l=

9 ? 22

? l = 1260 cm

Class Interval x f

0-20

10 7

20-40 30 P

40-60 50 10

60-80 70 9

80-100 90 13

xf 70 30P 500 630 1170

Since OR || BA (by construction)

39+P 2370+30P

?OPA + ?POR = 180? ?POR = 180? ? 90? = 90? Similarly ?QOR = 90? ?POR + ?QOR = 180?

2370 + 30P = 54 39 + P

2370 + 30P = 2106 + 54P 264 = 24P

? POQ is straight line through O. So PQ

P = 11

is diameter.

6. a2b2x2 + b2x ? a2x ? 1 = 0

OR

or b2x(a2x + 1) ? 1(a2x + 1) = 0

PT = PS ? ?PTS = ?PST

or (b2x ? 1)(a2x + 1) = 0

But ?PTS + ?PST + 60? = 180? ?

i.e., (b2x ? 1) = 0 or (a2x + 1) = 0

2?PTS = 120? ? ?PTS = ?PST = 60? \ PTS is an equilateral triangle \ TS = PT = 5 cm

1

1

So, x = b2 or x = ? a2

1

1

\ The roots of the equation are b2 and ? a2 .

Your Hard Work Leads to Strong Foundation

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PRE-NURTURE & CAREER FOUNDATION DIVISION

MATHEMATICS

SECTION-B

9. Lifetime Frequency

0-20

10

7. Class Interval f

c.f

15 ? 25

88

20-40

35

40-60

52

60-80

61

25 ? 35

10 18

80-100

38

35 ? 45

x 18 + x

100-120

29

45 ? 55

25 43 + x

Total

225

55 ? 65 65 ? 75 75 ? 85 85 ? 95

40 83 + x y 83 + x + y 15 98 + x + y 7 105 + x + y

Modal class is 60-80 61 - 52

Mode = 60 + 2 ? 61 - 52 - 38 ? 20

Total

140

=

60

+

9 122 - 90

? 20

Median = 58

8.

ALLEN Median class : 55 ? 65

N - c.f.

Median = l + 2

?h

f

70 - 43 - x

58 = 55 +

40

? 10

3 = 70 - 43 - x 4

12 = 27 ? x x = 27 ? 12 = 15 y = 20

90?

B

=

60

+

9 ? 20 32

45 = 60 + 8 = 60 + 5.625 = 65.625

10. Let AC be the cliff and ED be the ship.

AC = (h + 12) m

Now, in rt DAEB

h tan 45? = EB tan 45? =1 ? h = EB In rt DEBC,

A

h

45?

...(1) E 30? B

12 m

12 m

O

BC

D

C

P

tan 30? = EB

3.5 cm

A

1 12 ? 3 = h [From (1) , h = EB]

Steps of construction

1. Draw circle with centre at O and radius 3.5 cm. 2. Construct radii OAand OB such that ?AOB =90?. 3. Draw perpendiculars to OA and OB at A

and B respectively and let they intersect at P. 4. Now, PA and PB is a pair of tangents inclined to each other at an angle of 90?.

? h = 12 3

\ Height of cliff = (12 + 12 3 )m.

= 12 (1 + 3 )m. OR

PQ = 50 metres is the height of the tower. Let AB = h metres be the height of the building. Angle of elevation of the top of the building from the foot of the tower = 30?, i.e., ?AQB = 30?.

2/4

Your Hard Work Leads to Strong Foundation

CLASS - X (CBSE) STANDARD

MATHEMATICS

(Top of Tower)

P

So, volume of the cylinder = pr2h 2

= pr2 ? 10 3 cm3

(Top of Building) A

h

50m

\

pr2

?

10

2 3

=

4 3

p(98)

? pr2 ?

32 3

=

4 3

p ? 98

60? B

30? Q

Angle of elevation of the top of the tower from

the foot of the building = 60?, i.e., ?PBQ = 60?.

49

7

? r2 = 4 ? r = 2 cm

\ Diameter of solid cylinder = 7 cm.

OR

ALLEN AB

In DAQB , BQ = tan30?

h=1 BQ 3

? BQ = h 3

..(i)

In DPBQ

50 = tan 60? = 3 BQ

? BQ = 50

..(ii)

3

From (i) and (ii), we have h 3 = 50

3

Radius of hemispherical tank =

Diameter 2

=

1 2

m

Volume of tank =

2 3

pr3

2 22 1 11 = 3 ? 7 ? 8 = 42 m3

Volume of water to be emptied

11 = 42 ? 1000 litres

11 Since water emptied per second is 7 litres, So,

11000 42 litres will be emptied in

11000 7 1000 42 ? 11 s = 6 seconds

?

h =

50 m, 3

i.e., h = 16 2 m

3

500 = 3 seconds = 166.6 seconds.

\ The heights of building is 16 2 m .

3

SECTION-C

11. Volume of metal in spherical shell

=

4 3

p[53

?

33]

cm3

=

4 3

p

?

(125

?

27)

cm3

=

4 3

p ? 98 cm3

Let r be the radius of solid cylinder.

12. BQ = 12 cm, OB = 13 cm

A

x

13

P

\ OQ = 132 - 122

y QO

B

= 160 -144 = 25

OQ = 5 cm

Let PQ = y and PA = x In D POA : x2 + 132 = (y + 5)2 x2 + 169 = y2 + 10y + 25

: x2 ? y2 + 169 ? 25 = 10y ..(1)

Your Hard Work Leads to Strong Foundation

3/4

PRE-NURTURE & CAREER FOUNDATION DIVISION

MATHEMATICS

In DPQA : x2 = 122 + y2

x2 ? y2 = 144

..(2)

Put (2) in (1) 144 + 169 ? 25 = 10y

10y = 288 ? y = 28.8

\N = 30 =15 22

Which lies in the cumulative frequency 22, whose median class interval is 0.08-0.12.

PA = x = 144 + (28.8)2 = 973.44

= 31.2 cm

13. (i) Let assumed mean, A = 0.10 Table for assumed mean method and their product

N - cf Median = l + 2 ? h

f

with corresponding frequency is given

below

=

0.08

+

15 -13 ? 0.04 = 0.0888 9

Concentration Frequency Class

ALLEN of

SO2 (in ppm)

(fi)

0.00-0.04

4

mark (xi) 0.02

di=xi?A ? 0.08

fidi ? 0.32

14. (i) Total distance covered by competitor in placing the second potato in bucket.

0.04-0.08 0.08-0.12

9

0.06 ? 0.04 ? 0.36

9

0.10=A

0

0

= 2 ? 6 + 2 ? (6+4)

0.12-0.16 0.16-0.20

2

0.14

0.04

0.08

4

0.18

0.08

0.32

= 32 m

0.20-0.24 Total

2 S fi=30

0.22

0.12

0.24

Sfidi = ? 0.04

(ii) The distance run by the competitor to

Here, total observations (Sf1) = 30 and Sf1d1 = ? 0.4

According to formula,

picks up the first potato, second potato, third potato, fourth potato.....are respectively 2x6 2x{6+4), 2x(6+4+ 4),

2x(6+4+ 4+4+4)... i.e. 12, 20, 28, 36.....

Mean = A +

Sf1d1 = 0.10 + (-0.04)

Sf1

30

Clearly, it is an AP with first term, a = 12 and common difference, d = 20 ?12 = 8

= 0.10 ? 0.001 = 0.099 ppm (ii) Here, N = 30

Concentration of SO2 Frequency Cummulative frequency

s10 =

10 [2 ?12 + 9 ?8] 2

(in ppm)

(f1)

(cf)

= 5 [24 + 72] = 5 ? 96

0.00-0.04

4

4

0.04-0.08

9

13

= 480 m

0.08-0.12

9

22

0.12-0.16

2

24

0.16-0.20

4

28

0.20-0.24

2

30

4/4

Your Hard Work Leads to Strong Foundation

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